Question

Two identical spheres with charge $Q$ exert a force $F$ on each other. The spheres are kept fixed with a distance $r$ between them. A third identical but uncharged sphere C is brought in contact with A and then placed at the mid point of the line joining A & B. The magnitude of net electric force on C is

• A. $F$
• B. $\frac{F}{4}$
• C. $\frac{F}{2}$
• D. $F$

Answer 1

The correct option is A $F$

$F=\frac{k{Q}^2}{r^2}\underset{Q}{\underset{\cdot }{A}}\underset{Q}{\underset{\cdot }{B}}$

When, uncharged sphere will touch charged sphere charges will distribute equally as they are identical, so as to maintain same potential. Therefore, charges on A and C $=\frac{Q}{2}$

$\underset{Q/2}{\underset{\cdot }{A}}\underset{Q/2}{\underset{\cdot }{C}}\underset{Q}{\underset{\cdot }{B}}$

$F_{onCbyA}=\frac{K\frac{Q}{2}.\frac{Q}{2}}{\frac{r}{2}.\frac{r}{2}}=\frac{K{Q}^2}{r^2}=F{F}_{onCbyB}=\frac{KQ.\frac{Q}{2}}{\frac{r}{2}.\frac{r}{2}}=2\frac{K{Q}^2}{r^2}=2F{F}_{net}=2F-F=F$